{"id":206,"date":"2015-03-12T17:51:22","date_gmt":"2015-03-12T17:51:22","guid":{"rendered":"http:\/\/www.billconnelly.net\/?p=206"},"modified":"2025-07-05T04:15:34","modified_gmt":"2025-07-05T03:15:34","slug":"a-cheap-lofi-speaker-impedance-ii","status":"publish","type":"post","link":"https:\/\/www.billconnelly.net\/?p=206","title":{"rendered":"A cheap LoFi speaker – Impedance II"},"content":{"rendered":"

I needed to make a little speaker. The real reason? Because of a retro gaming session at work. However, there have been several times that I’ve needed a small cheap speaker that was suitable for delivering simple stimuli. I thought I would post a schematic and take this opportunity to give you an example of how Matlab can make impedance calculations easier, something I’ve talked a little about before<\/a>.<\/p>\n

\"speaker\"<\/a><\/p>\n

<\/p>\n

The core of this amplifier is an LM1875<\/a> om-amp. The circuit shown above is a complete rip off of the schematic shown in Figure 2 of the data sheet, with a few unnecessary parts removed. Instead of trying to design a one from scratch, the point of this post will be to analyse the schematic given. After that, you’ll hopefully be able to make any changes.<\/p>\n

Because we want to use a single sided power supply (i.e. our power supply gives us a positive rail and ground), we will need to “bias” the input to our amplifier, so when it receives no other input, it rests at half the supply voltage, giving the op-amp the ability to move negatively, and positively. We then place a capacitor in between the audio source (Vin<\/sub>) and the biased input, so that the bias voltage doesn’t leak back into the audio out. We also need some capacitors to stabilize our supply voltage, just in case the op amp needs to draw lots of current. All of this combines to be the input stage of our amplifier. We then want to amplify our input in a vaguely linear fashion, so we have negative feedback (where we can try to make sure the amplifier doesn’t waste current amplifying our bias voltage). Finally, on the output we have a large capacitor to make sure the speaker doesn’t see any of the bias voltage. The input stage is a bit of a pain to analyse, but I’ll be gentle.<\/p>\n

Input Stage<\/strong><\/p>\n

All these capacitors and resistors will end up creating quite a filter, and we need to be aware of what properties it will have. To do a full analysis of this with pen and paper would be a real pain, but with Matlab, this will be quite quick.<\/p>\n

\"singleopamp-input-raw\"<\/a>So the input stage is the parts of the circuit shown on the right. It looks complex to deal with because on top of all the capacitors, it also has a second voltage source. However, because at this point in the circuit we have solely linear elements, we can use a a technique called “superposition<\/a>“. In this, what we do is analyse the circuit twice, once for each voltage source, and then add the results. In turn, we change all voltage sources to short circuits, apart from one. Thus, if we change Vcc to a short (to earth), not only does it get rid of the issue of the secondary voltage, but it also removes the 100 \u00b5F capacitor from the circuit (because it has an earth on both terminal).<\/p>\n

\"singleopamp-input-simple1\"<\/a> If we then redraw the input stage circuit, we can quickly see that the circuit wont be too difficult to solve. Let’s start with the two 22 k\u03a9 resistors, and one 10 \u00b5F capacitor in parallel, these can be replaced with a single impedance $latex Z_{p} $. We know that the impedance of a capacitor is given by:<\/p>\n

$latex Z_{cap} = \\frac{1}{j \\omega C} &s=1$<\/div>\n

and for impedances in parallel, the equivalent impedance is:
\n\"singleopamp-input-simple2\"<\/a><\/p>\n

$latex Z_{equiv} = \\frac{1}{ \\frac{1}{Z_{1}} + \\frac{1}{Z_{2}} + \\frac{1}{Z_{3}}} &s=1$<\/div>\n

Thus we can calculate the impedance of the three elements in parallel as:<\/p>\n

$latex Z_{p} = \\frac{1}{ \\frac{2}{22000} + j \\omega 10^{-6} } &s=1$<\/div>\n

In matlab we can then calculate $latex Z_{p} $.<\/p>\n

\r\n>> f = 0.1:0.1:100000; % Frequecy\r\n>> w = f*2*pi; %Convert to angular frequency\r\n>> zp = 1 ./ (2/22e3+j*w*100e-6);\r\n<\/pre>\n
<\/p>\n
\"singleopamp-input-simple3b\"<\/a>We can then calculate the impedance of the whole right hand side ( $latex Z_{R} $ ), by adding $latex Z_{p} $ to the 10 \u00b5F capacitor and 22 k\u03a9 resistor that are in series with it.<\/p>\n
\r\n>> zr = zp + 22e3 + 1./(j*w*10e-6);\r\n<\/pre>\n
At this point, we should look back at the simplified circuit of the input stage. We can see that the voltage $latex V_{+} $ is simply $latex V_{in} $ minus the voltage lost over the 1 \u00b5F capacitor $latex V_{cap} $. In order to calculate that, we need to know that current which passes through the  $latex Z_{R} $ ( $latex i_{R} $), which is simply found via Ohm’s law.<\/p>\n
$latex  i_{R} = \\frac{ V_{in} }{ Z_{R} } &s=1$<\/div>\n
$latex  V_{cap} = i_{R} \\frac{1}{j \\omega 1^{-6}} &s=1$<\/div>\n
$latex  V_{+} = V_{in} – V_{cap}  &s=1$<\/div>\n
$latex  V_{+} = V_{in} – i_{R} \\frac{1}{j \\omega 1^{-6}} &s=1$<\/div>\n
$latex  V_{+} = V_{in} – \\frac{ V_{in} }{ Z_{R} } \\frac{1}{j \\omega 1^{-6}} &s=1$<\/div>\n
Factoring…<\/p>\n
$latex  V_{+} = V_{in} (1 – \\frac{ 1 }{ Z_{R} } \\frac{1}{j \\omega 1^{-6}} ) &s=1$<\/div>\n
Finally, the transfer function is…<\/p>\n
$latex  \\frac { V_{+} } { V_{in} } =  1 – \\frac{ 1 }{ Z_{R} } \\frac{1}{j \\omega 1^{-6}} &s=1$<\/div>\n
Which we can calculate and plot in Matlab.<\/p>\n
\r\n>> input_transfer = 1 - 1 ./ zr * 1./(j*w*1e-6);\r\n>> loglog(f, abs(input_transfer)); xlabel('Frequency (Hz)'); ylabel('|V(out)/V(in)|');\r\n<\/pre>\n
\"Input-transferFunction\"<\/a> So, we have made a high pass filter, with a cut-off of about 10 Hz. But remember, we’re trying to do a full analysis here, and we’re using superposition, so now we have to turn $latex V_{in} $ into a short to ground, and repeat our analysis. But don’t freak out, this is going to be super simple. Why? Because we can treat $latex V_{CC} $ as being purely DC, and hence all of the capacitors will act as open circuits.<\/p>\n
So if we convert all the capacitors to open circuits, we are left with a circuit with three resistors, one of which isn’t in the circuit because it is in series with a capacitor. Now the circuit has been simplified to a voltage divider with equal resistors. We know that means the input voltage ($latex V_{CC} $) will be halved, and so in this case, <\/p>\n
$latex  V_{+} = V_{CC}\/2 $.<\/div>\n
To complete the analysis, we add the two cases together, and so:<\/p>\n
$latex  V_{+} = V_{in} (1 – \\frac{ 1 }{ Z_vv{R} } \\frac{1}{j \\omega 1^{-6}} ) +  V_{CC}\/2   &s=1$ <\/div>\n
Feedback Stage<\/strong><\/p>\n
\"singleopamp-feedback\"<\/a><\/p>\n
This wont be as bad as the above. We just need to remember that because we have negative feedback, the voltage at the inverting terminal of the op-amp $latex  V_{-} $ will be the same as at the non-inverting terminal. We also need to use superpositioning again.<\/p>\n
If we look from the perspective of $latex  V_{out} $, then $latex  V_{-} $ looks like the center of a voltage divider, between a resistor, and a resistor\/capacitor. First, let’s calculate their impedance. The feedback resistor ( $latex  R_{f} $ ) has a simple impedance of 22 k\u03a9. The resistor\/capacitor section ( $latex  Z_{g} $) has a complex impedance. So in matlab:<\/p>\n
\r\n>> rf = 22e3;\r\n>> zg = 22e3 + 1./(j*w*10e-6);\r\n<\/pre>\n
The point of the voltage divider is, unsurprisingly to divide the voltage. Specifically, as $latex R_{f} $ gets larger relative to $latex  Z_{g} $ then $latex  V_{out} $ has to get larger and larger to keep $latex  V_{-} $ equal to $latex  V_{+} $ (because more of the voltage is generated over $latex R_{f} $ rather than $latex  Z_{g} $). With a normal resistive voltage divider, we could say $latex v_{out} = \\frac{v_{in} R_{2} }{R_{1} + R_{2}} $, here we can say:<\/p>\n
$latex  V_{-} = \\frac{ V_{out} Z_{g} }{ R_{f} + Z_{g}}  &s=1$ <\/div>\n
However, as we know the value of $latex  V_{-} $, we can rearrange, and get:<\/p>\n
$latex V_{out} = \\frac{V_{-} ( R_{f} + Z_{g} )}{ Z_{g}} &s=1$ <\/div>\n
So, given that we know $latex  V_{-} $ is equal to $latex  V_{+} $ we can now say (for the superposition case where $latex V_{CC} $ is off):<\/p>\n
$latex V_{out} = \\frac{ V_{in} (1 – \\frac{ 1 }{ Z_{R} } \\frac{1}{j \\omega 1e-6} ) ( R_{f} + Z_{g} )}{ Z_{g}} &s=1$ <\/div>\n
$latex \\frac{V_{out}}{V_{in}} = \\frac{  (1 – \\frac{ 1 }{ Z_{R} } \\frac{1}{j \\omega 1^{-6}} ) ( R_{f} + Z_{g} )}{ Z_{g}} &s=1$ <\/div>\n
So in Matlab, once we notice that most of that equation is the input_transfer<\/code> variable we calculated earlier:<\/p>\n
\r\n>> full_transfer = input_transfer .* (rf+zg) ./ zg;\r\n<\/pre>\n
Here it would be a good idea to think about what is actually going on. We did some filtering at the input stage, and then applied some amplification and filtering. We can view those all by plotting:<\/p>\n
\r\n>> loglog(f, [abs(transfer); abs((rf+zg) ./ zg); abs(full_transfer)]); xlabel('Frequency (Hz)'); ylabel('|V(out)/V(in)|');\r\n>> legend('Input Transfer Function', 'Feedback Transfer Function', 'Full Transfer Function')\r\n<\/pre>\n
\"Input-transferFunction-full\"<\/a><\/p>\n
So as you can see, most of the high pass filtering was achieved by the passive input filter, but because of the series capacitor to ground in the feedback path, low frequency signals were not amplified (because to DC signals, the ground half of the voltage divider provided infinite impedance).<\/p>\n
Finally, we should address the other superposition case, where there is no input apart from $latex V_{CC} $. We know that in this case, $latex V_{+} = V_{CC}\/2 $. Because of negative feedback, that means $latex V_{-} = V_{CC}\/2 $. Using the equation we calculated earlier, we can then say:<\/p>\n
$latex V_{out} = \\frac{V_{CC}\/2 * ( R_{f} + Z_{g} )}{ Z_{g}} &s=1$ <\/div>\n
However, here we need to stop and think. $latex V_{CC} $ is purely DC. That means all the capacitors act as open circuits. That means $latex  Z_{g} \\Rightarrow  \\infty  $. Which means that the above equation is really:<\/p>\n
$latex V_{out} = V_{CC}\/2 &s=1$ <\/div>\n
We could have guessed this from looked at the graph above, and the feedback transfer function: it tends towards 1 for DC values.<\/p>\n
Thus, the true, absolute final output equation is:<\/p>\n
$latex V_{out} = \\frac{ V_{in} (1 – \\frac{ 1 }{ Z_{R} } \\frac{1}{j \\omega 1e-6} ) ( R_{f} + Z_{g} )}{ Z_{g}} + V_{CC}\/2 &s=1$ <\/div>\n
.<\/p>\n
Output Stage<\/strong><\/p>\n
There is no point doing a full output analysis here, because the speaker is so cheap that it doesn’t have full specs. The only thing worth noting is that the huge capacitor on the output blocks the DC component of the output, i.e. $latex V_{CC}\/2 $. By being such a large capacitor, it presents almost no impedance to anything but the lowest of frequencies.<\/p>\n
 Build <\/strong><\/p>\n
Pallet wood, a 2.25″ hole and some screws<\/p>\n\n\n\n\n\t\n\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t
Description<\/th>Supplier<\/th>Part Number<\/th>Cost<\/th>\n<\/tr>\n<\/thead>\n
LM1875 Op-amp<\/td>Farnell \/ Newark<\/td>1468913 \/ 41K3523<\/td>\u00a32.05 \/ $3.08<\/td>\n<\/tr>\n
8 Ohm speaker<\/td>Farnell \/ Newark<\/td>4262323 \/ 19J5689<\/td>\u00a31 \/ $3.33<\/td>\n<\/tr>\n
 2200 uF, Electrolytic Capacitor<\/td>Farnell \/ Newark<\/td>2282194  \/ 65R2740<\/td>\u00a30.245 \/ $0.198<\/td>\n<\/tr>\n
10 x 100 uF, Electrolytic capacitors<\/td>Farnell \/ Newark<\/td>1902882 \/ 55T0367<\/td>\u00a30.34 \/ $0.36<\/td>\n<\/tr>\n
10 x 10 uF Electrolytic capacitor<\/td>Farnell \/ Newark<\/td>1902913 \/ 55T0590<\/td>\u00a30.685 \/ $0.39<\/td>\n<\/tr>\n
10 x 22k Resistors<\/td>Farnell \/ Newark<\/td><\/td>\u00a30.1 \/ $0.1<\/td>\n<\/tr>\n
12 V DC supply<\/td>Farnell \/ Newark<\/td>2424065 \/ 83T4338<\/td>\u00a35.41 \/ $10.04<\/td>\n<\/tr>\n
<\/td><\/td>TOTAL<\/td>\u00a39.83 \/ $17.50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n
\"singleopamp-back\"<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"
I needed to make a little speaker. The real reason? Because of a retro gaming session at work. However, there have been several times that I’ve needed a small cheap speaker that was suitable for delivering simple stimuli. I thought I would post a schematic and take this opportunity to give you an example of…Read more A cheap LoFi speaker – Impedance II<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[2,7],"tags":[],"_links":{"self":[{"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=\/wp\/v2\/posts\/206"}],"collection":[{"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=206"}],"version-history":[{"count":23,"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=\/wp\/v2\/posts\/206\/revisions"}],"predecessor-version":[{"id":244,"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=\/wp\/v2\/posts\/206\/revisions\/244"}],"wp:attachment":[{"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.billconnelly.net\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

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