transistor basics ( used as buffer ) - Page 1
Forum section: electronics / beginners

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 10:33:11 am
content: Hello, i had a question about the circuit below, basically it shows the VE is equal to 8.3V and i understood that, because VE=VB-0.7V (diode drop)but why the signal goes from 0 to 8.3V and not from 0.7V to 8.3V? can someone explain? there is parasitic capacitor that cancels the dc component or where is my misunderstanding? thanks.

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2025, 11:04:43 am
content: It would help if you posted a higher resolution schematic.The signal goes to 0V because the base voltage is zero and the emitter is pulled down by the 1k resistor.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 11:06:52 am
content: q

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 11:10:23 am
content: I dont know why it keeps such quality, base voltage isnt 0 tho, as vin there is a sinusoidal wave that goes from 0.7V to9, my new theory is that until it doesnt reach the 0.7v the diode doesnt conduct so we wont see anything in output, until it reaches 0.7v , since it reaches that value u can see as output the wave which is lowered by 0.7v cus of diode drop. ( tho im not sure if someone could confirm my hypothesis i would be thankful )

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2025, 11:12:06 am
content: Those images are all too small. 120x72 resolution is too low.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 11:20:00 am
content: https://sites.google.com/site/roelarits/home/articles/the-transistor/transistor-as-a-bufferhere, u can see it

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2025, 11:42:28 am
content: Quote from: fady232 on April 15, 2025, 11:20:00 amThat site gives me a warning "Potential Security Risk Ahead" so I'm not going to visit it.Please attach the schematics with a high enough resolution to be able to read.

author: Sorama
author_rank: Frequent Contributor
timestamp: April 15, 2025, 11:50:44 am
content: Quote from: Zero999 on April 15, 2025, 11:42:28 amI think your browser is telling you nonsense.The images are low res, nothing OP can do about it.

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2025, 11:53:10 am
content: Quote from: Sorama on April 15, 2025, 11:50:44 amI tried it in Edge and Firefox. Perhaps, it's the company firewall?Is it the OP's website? Either way, there's not much we can do, without clear schematics.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 12:02:24 pm
content: here u go

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2025, 12:07:13 pm
content: Quote from: fady232 on April 15, 2025, 12:02:24 pmThat's much better thank you. The results are as one should expect: the emitter voltage follows, the base, less about 700mV.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 12:13:57 pm
content: So what happens in the circuit is that i got a sine wave from 0.7v to 9v, and by having a drop of 0.7V is like shifting the wave down by 0.7V right?. So its incorrect if i say that there will be a sine wave that starts from 0V in output because until 0.7V the diode wont conduct, and it will go till 8.3V, because 0.7v drop will be on the diode?

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2025, 12:16:27 pm
content: Quote from: fady232 on April 15, 2025, 12:13:57 pmYes, but not the base-emitter voltage won't be exactly 0lV and will vary a little, depending on the voltage.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 12:36:08 pm
content: kk i understood, thanks

author: MrAl
author_rank: Super Contributor
timestamp: April 15, 2025, 01:13:22 pm
content: Here is the original diagram.The output will almost follow the input so they call it a voltage follower. What is not immediately apparent is that there will be a small deviation due to the way the base emitter diode responds with the 1k load. As the input voltage increases the base emitter diode will drop an additional voltage that is a small percentage of that 0.7 volts. That would cause a little distortion to the input sine wave, but it should not be too much. If the 1k was increased in value, the distortion would be even less.As the input gets closer to 9v (in this schematic) the distortion will increase though.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 01:29:20 pm
content: why a higher resistor causes less distortion? can u explain me?

author: PGPG
author_rank: Frequent Contributor
timestamp: April 15, 2025, 01:35:58 pm
content: Quote from: fady232 on April 15, 2025, 10:33:11 amTrully it is hard to understand where your problem comes from.At input you have signal going from 0.7V to 9V.At output you have signal all the time 0.7V below it.So for 0.7V at input you should expect 0.7V-0.7V=0V at output.For 9V at input you should expect 9V-0.7V = 8.3V.So output voltage varies from 0V to 8.3V and it comes from what you seem to know (you have written "and i understood that, because VE=VB-0.7V (diode drop)").So where from your idea of getting at output 0.7V to 8.3V.Input never drops below 0.7V so the 0.7V is 'reached' all the time.

author: PGPG
author_rank: Frequent Contributor
timestamp: April 15, 2025, 01:40:34 pm
content: Quote from: fady232 on April 15, 2025, 01:29:20 pmWhen input voltage rises the transistors drives the output high and can give at output the current that is needed.When input voltage falls the only what can drive the output down is the resistor. The higher resistor the lower current it can issue and it is easier that it is not enough to take output down enough fast.Specially near 0V at output as when coming to 0V at output the current in resistors reaches 0mA.

author: fady232
author_rank: Contributor
timestamp: April 15, 2025, 01:52:48 pm
content: so by selecting a higher resistor, it reaches 0V ''slowly'' and the sine wave isnt too much distorted, right?

author: MrAl
author_rank: Super Contributor
timestamp: April 15, 2025, 02:09:40 pm
content: Quote from: fady232 on April 15, 2025, 01:29:20 pmHi,Sure.The diode current and voltage curve is not linear it is slightly nonlinear even with low current changes.For example, and this is just an example, if the currents we measured were 1ua, 2ua, 4ua, and 8ua, we might see the diode drop voltages of:0.567v, 0.588v, 0.609v, 0.629v respectively.That means that the output will change slightly less than if the diode drop was always 0.7 volts. That means that the emitter does not follow the base voltage exactly.It could be close enough for some applications, but for high quality audio for example it may not be acceptable.If we decrease the value of the emitter resistor, the current would be changing even more and thus would create more distortion. Imagine if the emitter resistor was zero Ohms, then we would see maximum distortion. With a resistor of 1 Megohms (if that was possible), we could see much less distortion because there would be hardly any current.When we apply an actual load it would get worse to some degree.We could look at some more exact numerical examples if you like.

author: mawyatt
author_rank: Super Contributor
timestamp: April 15, 2025, 02:45:08 pm
content: If the OP is concerned about output offset and/or distortion a Complementary EF can help. With a Complementary EF (within the linear range) the output becomes:Vout ~ Vin -Vbenpn + Vbepnpand since |Vbe npn| ~ |Vbepnp| then:Vout ~ VinHere's a simple LTspice circuit and results with Vsupply 9V and Vin from 0 to 10V;Edit: You can change the input range to 1 to 9V and get the results in the last two plots.Hopefully this helps.Best

author: fady232
author_rank: Contributor
timestamp: April 18, 2025, 07:06:35 am
content: ye i saw this kind of configuration to compensate